Since the average acceleration is along Δv (a=Δv/Δt), the average acceleration is perpendicular to Δr. We already know that, since the path is circular, v is perpendicular to r and so is v' to r', according to the figure given by you. (Since the velocity vectors v and v' are always perpendicular to the position vectors, the angle between them is also ΔΘ).
Note that the book stated that If we place Δv on the line that bisects the angle between r and r', we see that it is directed towards the centre of the circle. It was just a verification of the statement already quoted above.
A perpendicular from the centre bisects the chord, and since PCP' is an isosceles triangle, bisects the angle between r and r' (from the angle bisector theorem). Obviously it will be directed towards the centre considering the geometry of the figure,
or understand it this way,
The perpendicular bisector of a chord passes through the centre of the circle. It is a very fundamental theorem which has many applications in various fields of Physics and Mathematics. I have attached a link to its proof .
Actually, it is not a rule, it is just an approach to introduce this topic at the beginner level, assuming that you know basic geometry, upto 10th grade.
In a nutshell, since Δv is perpendicular to Δr, it passes through the centre of the circle.
And so, the average acceleration is directed towards the centre.
Note that the average acceleration only changes into instantaneous acceleration if we put the limits (Δt->0). So, its direction is towards the centre. In the fig(c), we are only approximating the situation on an infinitesimally small scale. But the overall concept remains the same.
P.S.: Read my comments on your question.